- #1

binbagsss

- 1,266

- 11

##U## has 3 dof. ##\phi## has 1.If asked why this works, in terms of a dof argument, why is this?e.g . compared to GR common decomposition of the Riemann tensor into##R^{\nu\lambda\kappa\rho} =R*^{\nu\lambda\kappa\rho} + BR\eta*^{\nu\lambda\kappa\rho} + C^{\nu\lambda\kappa\rho},## (1)

With ##C^{\nu\lambda\kappa\rho}## is traceless, ##R=\eta_{\nu\kappa} \eta_{\lambda\rho}R^{\nu\lambda\kappa\rho}##, where #\eta^{ab}# denotes the Minkowski metric and ##B,C## are constants, and where ##R*^{\nu\lambda\kappa\rho}## denotes ##R*^{\nu\lambda\kappa\rho}=\frac{1}{4}\left[{R}^{\nu\kappa}\eta^{\lambda\rho}-R^{\nu\rho}\eta^{\lambda\kappa}+{R}^{\lambda\rho}\eta^{\nu\kappa}-{R}^{\lambda\kappa}\eta^{\nu\rho}\right] ## and## \eta*^{\nu\lambda\kappa\rho}= 2\eta^{\nu\kappa}\eta^{\lambda\rho}-2\eta^{\nu\rho}\eta^{\lambda\kappa} ##

From what I understand the tensors making up the decomposition can be a number of degrees of freedom at most the same as the object they are making up, and altogether they must have at least the total number of dof of the object they are making up. So , e.g. if T=A+B+C, and T has 10 d.o.f, then each of A,B,C can have any number between [1,10] of d.o.f, and altogether they must have at least 10 d.o.f? Is this correct?

So looking at the above, equation (1), the first term has 10 dof, the second 1, and the last 10 dof. Or would you look at the first term as a sum of four terms each of 10 dof, so it could potentially represent a tensor with 40 dof?

Thanks.